Todays cryp question ..

PDA
*HAD
-----------
NHHZ
DNZE
DIED
---------
ANDAAZ
------------


What is 5(H-A) ?

• 1) As A*A=E so value of A can't be 5 & 6.
  2) As well A=D+1. So value of D can't be 4 and 5.
  3) Assume D=2, then A=3. We get H=4.
  4) Value of P must be 7, bcz we got H=4. and 2*7=14.
  this gives N=1.
  Now with these value we can calculate rest of the part.
  
  Solution is
  
  7 2 3
  4 3 2
  -----
  1 4 4 6
  2 1 6 9
  2 8 9 2
  ------------
  3 1 2 3 3 6
  ------------
  
  
  Answer of 5(H-A)= 5(4-3)
  = 5 * 1 =5.
• 10 years agoHelpfull: Yes(10) No(0)
• i got the hint from the last question asked: i.e H+A+N+D = ?
  Options: 7, 8, 9, 10.
  Therefore, its true that the sum of H+A+N+D will be less than or equal to 10.
  soln:
  -the first letter of ANDAAZ : A=D+1 ----------------------------(eq-1)
  - then A*D = Z nd D*D=H or D*D +1 = H
  - then in PDA * A : A*A = E nd D*A = Z or D*A +1 = Z
  so A can be only 2 or 3 then E= 4 or 9
  - from eq-1 : D+1 = A i.e D can be either 1 or 2 if A = 2 or 3.
  - on solving for the two possible values we can the solution:
  
  723
  *432
  ---------
  1446
  2169
  2892
  --------
  312336
  ---------
  so, H+A+N+D=10
  and 5(H-A)= 5(4-3)=5.
• 10 years agoHelpfull: Yes(7) No(3)
• P D A
  * H A D
  -------------
  N H H Z
  D N Z E _
  D I E D _ _
  -----------------
  A N D A A Z
  
  P D A ---(i)---
  * D
  -------
  N H H Z
  
  P D A ---(ii)---
  * A
  -------
  D N Z E
  
  P D A ---(iii)---
  * H
  -------
  D I E D
  
  From final product ANDAAZ it is clear that D+1=A.
  From (i) & (ii) ---> A x D = Z and Z less than 10.
  From (ii) ---> Carry over D is 0, hence A x A = E and E less than 10.
  Now A can be 1, 2 or 3.
  A cannot be 1 as A x D = Z.
  Taking A=2, then E=4, D=1 (since, D+1=A), Z=2. But Z cannot be equal to 2. Hence A cannot be equal to 2.
  Taking A=3, then E=9, D=2, Z=6.
  From (i) ---> D x D = H or H=4 And D x P = NH or P=7 and N=1.
  Hence the result follows :::
  
  7 2 3
  * 4 3 2
  --------
  1 4 4 6
  2 1 6 9 _
  2 8 9 2 _ _
  ------------
  3 1 2 3 3 6
• 10 years agoHelpfull: Yes(3) No(0)
• i got the hint from the last question asked: i.e H+A+N+D = ?
  Options: 7, 8, 9, 10.
  Therefore, its true that the sum of H+A+N+D will be less than or equal to 10.
  soln:
  -the first letter of ANDAAZ : A=D+1 (eq-1)
  - then A*D = Z nd D*D=H or D*D +1 = H
  - then in PDA * A : A*A = E nd D*A = Z or D*A +1 = Z
  so A can be only 2 or 3 then E= 4 or 9
  - from eq-1 : D+1 = A i.e D can be either 2 or 8 but * not possible since H+A+N+D
• 10 years agoHelpfull: Yes(1) No(0)
• P D A
  * H A D
  -------------
  N H H Z
  D N Z E _
  D I E D _ _
  -----------------
  A N D A A Z
  
  P D A ---(i)---
  * D
  -------
  N H H Z
  
  P D A ---(ii)---
  * A
  -------
  D N Z E
  
  P D A ---(iii)---
  * H
  -------
  D I E D
  
  From final product ANDAAZ it is clear that D+1=A.
  From (i) & (ii) ---> A x D = Z & Z Carry over D is 0, hence A x A = E D x D = H => H=4 And D x P = NH => P=7 and N=1.
  Hence the result follows :::
  
  7 2 3
  * 4 3 2
  --------
  1 4 4 6
  2 1 6 9 _
  2 8 9 2 _ _
  ------------
  3 1 2 3 3 6
• 10 years agoHelpfull: Yes(1) No(0)
• 723*
  432
  1446
  2169
  2892
  312336
  Firstly, d+1=a,a=3and d=2 suits best in the equation PDA*D =NHHZ. after that would be solved.as soon.as u get the value of a and d.
• 10 years agoHelpfull: Yes(1) No(0)
• easiest cryptarithmetic problem..........solved in first attempt
  
  723
  *432
  ------
  1446
  2169-
  2892--
  ------
  312336
  
• 10 years agoHelpfull: Yes(1) No(0)
• some improvement to dhanjay lumar's solution
  as D= 1 or 2, now D=1 is not possible because if D=1 than 1*A= A ,so unit digit would be A instead of Z. hence D=2 onlyyy.
• 10 years agoHelpfull: Yes(0) No(0)
• PDA*A=DNZE, so possible E values = 1,4,9,6
  Corresponding poss A values= 9,(2,8),(3,7),4
  Now D=A-1;
  aand H+E=A+10/A
  considering H+E=A+10 (which usually happens)
  H=(A+10)-E
  = 8,4,8 (slight calculation is required eg- 9+10=19;19-1=18, not possible for value of H so ruled out)
  so H= 8,4,8
  possible CORRESPONDING A values 2,3,4
  possible CORRESPONDING E values 4,9,6
  now if A = 2, D=1 NOT POSSIBLE
  if A=3,D=2, E=9 and H=4
  putting the values in PDA*D=NHHZ, Z=6 & N=1 & P=7
  thereby other values can be easily sorted out.
• 10 years agoHelpfull: Yes(0) No(1)
• pda*had= 723*432
  1446
  2169
  2892
  ------
  312336
• 10 years agoHelpfull: Yes(0) No(0)
• P=7,D=2,A=3,H=4,N=1,I=8,E=9,Z=6 These value get by solving 723*432
  ---------
  1446
  2169*
  2892*
  -------
  312336
  --------
  ANDAAZ
  
  
• 10 years agoHelpfull: Yes(0) No(1)
• 723
  432
  .............
  1446
  2169
  2892
  ...............
  312336
  
  5(H-A)=5
• 10 years agoHelpfull: Yes(0) No(0)
• 7 2 3
  4 3 2
  ---------------
  1 4 4 6
  2 1 6 9
  2 8 9 2
  ------------------
  3 1 2 3 3 6
  -------------------
  
  
  
  5( H- A ) = 5 ( 4 - 3 ) = 5 * 1 = 5
• 10 years agoHelpfull: Yes(0) No(0)
• in this question see the last column from end : D + carry=A;
  it means D and A will be consecutive nos. and A will b greater than D. D and A can not be 1( we can see it from its product value).now try with A=3 and D=2; here in the first attempt you will reach to the solution.
  
  Ans:- 5
• 10 years agoHelpfull: Yes(0) No(0)